Dynamics of Structures: Theory and Analysis

Transcription

1 1. Free vibrations 2. Forced vibrations 3. Transient response 4. Damping mechanisms Dynamics of Structures: Theory and Analysis Steen Krenk Technical University of Denmark 5. Modal analysis I: Basic idea and matrix formulation 6. Modal analysis II: Implementation and system reduction 7. Damping and tuned mass dampers 8. Time integration by Newmark methods 9. Structural response to earthquakes 10. Vibration of cables, bars, etc. 11. Vibration of beams 12. Finite element formulation for bars, beams etc. 13. Course summary

2 Books Daniel J. Inman, Engineering Vibration, 2nd Edition, Prentice Hall International, Upper Saddle River, N.J., J.W. Tedesco, W.G. McDougal and C.A. Ross, Structural Dynamics, Addison- Wesley, Menlo Park, Ca, A.K. Chopra, Dynamics of Structures. Theory and application to Earthquake Engineering, Prentice-Hall, Upper Saddle River, N.J., Geradin, M. and Rixen, D., Mechanical Vibrations, Theory and Applications to Structural Dynamics, 2 nd ed., Wiley, Chichester, I. Langen and R. Sigbjørnsson, Dynamisk Analyse av Konstruktioner, Tapir, Trondheim, Journals Earthquake Engineering and Structural Dynamics. Journal of Sound and Vibration Journal of Engineering Structures. Journal of Structural Engineering. Journal of Engineering Mechanics. International Journal for Numerical Methods in Engineering. Computers and Structures.

3 General Introduction Offshore platforms Tyra South STAR platform Harald Field (

4 Wind Turbines Nysted Wind Farm (

5 Cable stayed bridge Erasmus Bridge, Rotterdam (

6 Rotating pedestrian bridge Gateshead Millennium Bridge, Gateshead (

7 Pedestrian bridge Solferino Bridge, Paris (S. Krenk)

8 Pedestrian suspension bridge Millennium Bridge, London ( ) See also the detailed presentation at

9 Lecture 1: Free Vibrations Undamped vibrations Damped vibrations Logarithmic decrement Energy balance State variables and state-space Discrete time increments

10 Undamped vibrations Equation of motion by balance of inertial force mẍ and elastic force kx with mass m and stiffness k, m ẍ + k x = 0 (1.1) Obtain normalized equation by division by m ẍ + ω 2 0 x = 0 (1.2) where the natural angular frequency ω 0 is defined by ω 0 = k m (1.3) Note that the (angular) frequency is determined by ω 2 0 = stiffness mass In many problems stiffness and mass can be estimated by use of energy methods.

11 Solution Solution requires initial conditions in terms of initial position x 0 and initial velocity v 0 = ẋ 0. x(t) = x 0 cos(ω 0 t) + v 0 ω 0 sin(ω 0 t) (1.4) x v / ω 0 ω 0 t ω 0 t Figure 1.1: a) displacement x(t), and b) velocity ẋ(t) as function of time t. The period T requires the phase ω 0 t to increase by 2π, Note that the (angular) frequency is determined by T = 2π ω 0 (1.5) The natural frequency f is the number of oscillations per time unit, whereby f = 1 T = ω 0 2π (1.6)

12 Damped vibrations Include energy dissipation through damping force cẋ, m ẍ(t) + c ẋ(t) + k x(t) = 0 (1.7) Note, that all three parameters m, c, k are positive, when the forces are restoring. Normalize equation by division by m ẍ + 2ζω 0 ẋ + ω 2 0 x = 0 (1.8) using the natural angular frequency ω 0, ω 0 = k m (1.9) and the damping ratio ζ, ζ = c 2 km (1.10) The characteristics of the free vibration behavior depends on the magnitude of the damping ratio ζ.

13 Solutions The free vibration solution is of exponential type and is found by substitution of the partial solution x(t) = Ae λt into the homogeneous equation of motion, yielding the characteristic equation λ 2 + 2ζω 0 λ + ω 2 0 = 0 (1.11) There are three different cases of damped vibrations, depending on the magnitude of the damping ratio ζ. The solutions are given below for initial conditions (x, ẋ) t=0 = (x 0, v 0 ). Underdamped, 0 < ζ < 1 : Two complex roots of characteristic equation, λ = ω 0 ( ζ ± i 1 ζ 2 ). Imaginary part is expressed by defining the damped natural frequency ω d = ω 0 1 ζ 2 (1.12) This leads to complex partial solutions of the form x(t) = A e ζω 0t }{{} amplitude e ±iω dt }{{} oscillation (1.13)

14 The last factor can be expressed in terms of sin(ω d t) and cos(ω d t). x(t) = x 0 e ζω 0t [ cos(ω d t) + ζω 0 ω d sin(ω d t) ] + v 0 ω d e ζω 0t sin(ω d t) (1.14) x v / ω 0 ω 0 t ω 0 t Figure 1.2: a) displacement x(t), and b) velocity ẋ(t) for ζ = The period of the oscillatory factor in damped free oscillations is T d = 2π/ω d.

15 Critically damped, ζ = 1 : The characteristic equation has the real double root, λ = ω 0, and in this case the amplitude A is replaced by a linear function A + Bt. x(t) = [ x 0 + (ω 0 x 0 + v 0 ) t ] e ω 0t (1.15) Overdamped, 1 < ζ : The characteristic equation has two real roots λ = ω 0 ( ζ ± ζ 2 1). Introducing the parameter ω d = ω 0 ζ2 1 < ζω 0 the solution is expressed in terms of hyperbolic functions as x(t) = x 0 e ζω 0t [ cosh(ω d t) + ζω 0 ω d sinh(ω d t) ] + v 0 ω d e ζω 0t sinh(ω d t) (1.16)

16 Logarithmic decrement x,a a j a j+1 ω 0 t Figure 1.3: Displacement record with maxima a j and a j+1 for ζ = Underdamping can also be characterized by the decrease in amplitude from one maximum to the next. The ratio between any two maximum values following each other is constant, and the logarithmic decrement is defined as ( ) aj δ = ln = ln ( ) e ζω 0T d 2π ζ = (1.17) 1 ζ 2 a j+1 or for lightly damped systems δ 2π ζ for ζ << 1 (1.18) NOTE: Lightly damped structures may have ζ as low as

17 Energy balance Form the rate of work by multiplication of the equation of motion with ẋ, ( ) ẋ(t) m ẍ(t) + c ẋ(t) + k x(t) = 0 (1.19) Rewrite as time derivative, ( ) d 1 dt 2 m ẋ k x2 = c ẋ 2 0 (1.20) This defines the mechanical energy E as E = E kin + E el (1.21) with kinetic energy and elastic energy E kin = 1 2 m ẋ2 (1.22) E el = 1 2 k x2 (1.23) The mechanical energy of undamped free vibrations is constant. When the motion can be described (estimated) by a single degree of freedom, the equation of undamped motion can be determined from the time derivative of the mechanical energy.

18 State vector and phase-plane Time histories of displacement x(t) and velocity v(t) = ẋ(t) against time t. x v / ω 0 ω 0 t ω 0 t Figure 1.4: a) displacement x(t), and b) velocity ẋ(t) as function of time t. Combination into a single three-dimensional graph as shown in Fig., with time t along the first axis. v / ω 0 v / ω 0 x ω 0 t x Figure 1.5: a) response path and b) phase-plane diagram. Projection of the state vector (x(t), ẋ(t)) on the phase-plane. Use of normalized coordinates x, ẋ/ω 0 reduces to near-circle.

19 Discrete time increment - Undamped system Undamped equation of free vibrations, ζ = 0, ẍ + ω 2 0 x = f(t) Free oscillations described entirely in terms of initial conditions, x = x 0 cos(ω 0 t) + ẋ 0 ω 1 0 sin(ω 0 t) ẋ = x 0 ω 0 sin(ω 0 t) + ẋ 0 cos(ω 0 t) State vector (x, ẋ) 1 at time t = t in terms of initial state vector (x, ẋ) 0. Recurrence relation for any pair of state vectors with time separation t. Time separation only appears in the form of the non-dimensional parameter α = ω 0 t Non-dimensional form, when ẋ is replaced with tẋ. [ ] [ ] [ x cos α α 1 sin α x = t ẋ α sin α cos α t ẋ i+1 ] i

20 Discrete time increment - Damped system Normalized equation for free vibrations ẍ + 2ζ ω 0 ẋ + ω0 2 x = 0 Damped natural angular frequency, ω d = ω 0 1 ζ 2 Free damped oscillations given in terms of the initial conditions ( x = x 0 e ζω 0t cos(ω d t) + ζω ) 0 sin(ω d t) + ẋ0 e ζω0t sin(ω d t) ω d ω d ẋ = x 0 ω 2 0 ω d e ζω 0t sin(ω d t) + ẋ 0 e ζω 0t ( cos(ω d t) ζω 0 ω d sin(ω d t) Two non-dimensional time scales, conveniently defined as α = ω d t, β = ζ ω 0 t ) Recurrence relations for time increment t, [ x ] t ẋ i+1 = e β α [ α cos α + β sin α sin α (α 2 + β 2 ) sin α α cos α β sin α ] [ x ] t ẋ i

21 Structure of time-stepping algorithm The exact solution (x n, ẋ n ) can be obtained at discrete times 0, t, 2 t, by starting at the initial conditions (x 0, ẋ 0 ) and multiplying with the matrix A, found above, in each step. y 0 = (x 0, t ẋ 0 ) T for i = 0 : n 1 y i+1 = A y i (x, t ẋ) = y T Table 1.1: Direct time-stepping algorithm. NOTE: The simplicity of the algorithm is obtained because the natural frequency and damping ratio is known. For multi-degree-of-freedom systems, this procedure requires a modal analysis, described later.

22 Exercise 1.1 The natural frequency of a single-degree-of-freedom-system depends on the ratio of stiffness to mass. For a simple mass spring system the relation is ω 2 0 = k/m. a) For the beam shown in the figure the force-displacement relation for a transverse force at the end is F = 3 EI l 3 What is the natural frequency for transverse vibrations of a heavy mass M fixed to the end of the beam, when the mass of the beam is neglected. b) Let the mass M = 1000 kg be supported by a beam of length l = 10 m. What is the bending stiffness EI necessary to give the frequency f = 2.0 Hz. u Figure 1.6: Beam supporting a mass M.

23 Exercise 1.2 Consider the mass m supported by a spring with stiffness k. The motion of the mass m is described by x. The mass of the spring m s is small relative to the mass m, and therefore the motion of the spring is quasi-static. This implies that the spring extends uniformly. Thus, the left end point is a rest, while the right end point moves x. A non-dimensional coordinate ξ is introduced such that the spring is described by 0 ξ 1. This implies that the motion of a point described by the position ξ is ξx. a) Find the total kinetic energy of the mass m and the spring in terms of ẋ. b) Use energy balance to find an expression for the natural angular frequency ω 0. c) The contribution from the spring can be included as an extra contribution to the effective mass m eff, ω 2 0 = k/m eff with m eff = m + (?)m s Find the coefficient (?) of the spring mass. Figure 1.7: Beam supporting a mass M.

24 Exercise 1.3 The figure shows a water column of total length l and cross-section area A. The mass density is ρ. When the water surface at the left side is displaced the distance x downward, the water surface at the right side is lifted the same distance x and conversely. The system is exposed to downward gravitation with acceleration constant g. a) Express the potential energy E pot and the kinetic energy E kin as function of x and ẋ, respectively. b) Use an energy balance argument to find an expression of the natural angular frequency ω 0. c) The angular frequency ω 0 is independent of the mass density ρ. Why? Figure 1.8: Beam supporting a mass M. The motion of the fluid may be constrained, whereby damping is introduced. This device can be used as damper of ship roll motion.

25 Summary Examples have been given of structures that often exhibit dynamic behavior and must be analyzed for dynamic effects. Free vibrations of a single degree-of-freedom system constitute an exchange between potential and kinetic energy. The time scale is characterized by the natural angular frequency ω 0. The square of the natural (angular) frequency is determined by the ratio stifness / mass, i.e. ω 2 0 = k/m. Damping is charactized by the non-dimensional damping ratio ζ, describing attenuation per vibration cycle. Vibrations with ζ < 1 are underdamped and may have damping ratio as low as Damping may be measured from attenuation of free vibration response in terms of the logarithmic decrement δ = 2πζ. The equation of motion may be considered as the time derivative of an energy balance equation. For simple systems with distributed mass or stiffness the equation of motion may be obtained from the energy balance relation. The displacement and velocity may be combined into a state vector (x, ẋ) describing the response of the system. Initial conditions and response are conveniently represented in a phase-plane.